Trigonometry: Solved Assignments and Essays

This page contains solved assignments and essays on Trigonometry. It includes solutions to questions on maximum distance, temperature, and tension in the string. The content covers Trigonometry 1 and other courses.

Trigonometry 1
Trigonometry
Student’s Name
Course
Professor’s Name
University
City (State)
Date

Trigonometry 2
Trigonometry
Question 1
Part i
x=6 sin 2 t−cos 2 t
Maximum distance of particle occurs when, d x
dt =0
d x
dt = d
dt ( 6 sin 2t−cos 2t )=12 cos 2 t +2 sin 2 t=0
12 cos 2 t=−2sin 2 t
sin 2t
cos 2 t =tan 2t=−6
2 t=tan−1 (−6)=1.7359
time , t=0.867971 seconds
when t=0.867971 , distance x=6 sin 1.7359−cos 1.7359=6.083 m
Part ii
The y-intercept occurs when, t=0 so that,
x=6 sin 0−cos 0=−1
x-intercepts occur when x=0 so that,
6 sin 2 t=cos 2 t
sin 2t
cos 2 t =tan 2t= 1
6

Trigonometry 3
2 t=tan−1
( 1
6 )=0.165149 , t=0.08257
Using the domain, 0 ≤ t ≤ 2.25 π
The maxima occurs at t=0.08257∧t=0.08257+ π=4.0096
The minima occurs at t=0.08257+ 0.5 π =2.4388∧t =2.4388+ π =5.5804
Using the intercepts, maxima and minima points, the plot of the function
x=6 sin 2 t−cos 2 t is shown in the figure below.

Trigonometry 4
Question 2
F ( t ) =60+10 sin π
12 (t−8) for 0 ≤ t ≤ 24
Part a
At 8:00am,
F ( 8 ) =60+10 sin π
12 ( 8−8 ) =60+10 sin 0=6 0 ℉
At 12:00pm,
F ( 12 )=60+10 sin π
12 ( 12−8 )=60+ 10sin ( π
3 )=6 8.66 ℉
Part b
F ( t )=60=60+10 sin π
12 (t−8)
10 sin π
12 (t−8)=60−60=0
sin π
12 (t−8)=0
π
12 ( t−8 ) =sin−1 0=0 , π , 2 π …
π
12 ( t−8 ) =0 ,t−8=0 , t=8 (8 :00 am)
π
12 ( t−8 ) =π , t−8=π × 12
π =12 , t=8+12=20(8 :00 pm )

Trigonometry 5
π
12 ( t−8 ) =2 π , t−8=2 π × 12
π =24 ,t =8+24=32(ignored)
Therefore, the temperature is 6 0 ℉ at 8am and 8pm.
Part c
Maximum temperature occurs when sin π
12 (t−8)=1
So that maximum temperature¿ 60+10 ( 1 )=70 ℉
So that, π
12 ( t−8 ) =sin−1 1= π
2
π
12 ( t−8 ) = π
2
t−8= π
2 × 12
π =6
t=8+ 6=14 (2 :00 pm)
Minimum temperature occurs when sin π
12 (t−8)=−1
So that minimum temperature¿ 60+10 (−1 )=50 ℉
So that, π
12 ( t−8 ) =sin−1 (−1)= 3 π
2
π
12 ( t−8 ) =3 π
2
t−8=3 π
2 × 12
π =18
t=8+18=26(2:00 am )

Trigonometry 6
Part d
t 0 2 8 10 14 20 24
x 51.34 50 60 65 70 60 51.34
Question 3
Part a
Tension=mgcosθ
If θ is very small, sinθ ≅ θ= x
L and cosθ=1−θ2
2
Tension=mgcosθ=mg (1−θ2
2 )=mg (1− ( x
L )2
2 )=mg (1−
x2
L2
2 )=mg (1− x2
2 L2 )

Trigonometry 7
Therefore, tension in the string will be mg (1− x2
2 L2 )
Part b
Restoring force ¿ mgsin θ
But, when θ is very small, sinθ≅ θ= x
L
Restoring force¿ mgsin θ=mg x
L
Part c
F=−mgsinθ where F is the restoring force.
From Newton’s second Law of Motion, F=ma so that,
F=ma=−mgsinθ=−mg x
l
ma=−mg x
l
a=−g
l x
Evidently, acceleration is directly proportional to the displacement x and in
the opposite direction.

Trigonometry 8